car going 15 m/s is brought to a rest in a distance of 2.0 m as it strikes a pile of dirt. how much average force is exerted by the seatbelts on a 90 kg passenger as the car is stopped?

f=5.1N

To calculate the average force exerted by the seatbelts on the passenger, we can use the equation:

Force (F) = (mass x change in velocity) / time

Given:
Initial velocity (u) = 15 m/s
Final velocity (v) = 0 m/s
Distance (s) = 2.0 m
Mass of passenger (m) = 90 kg

First, we need to calculate the time it took for the car to stop. We can use the equation:

s = (u + v) / 2 * t

Rearranging the equation to solve for time, we have:

t = (2s) / (u + v)

Substituting the given values:

t = (2 x 2.0 m) / (15 m/s + 0 m/s)
t = 0.27 seconds (rounded to two decimal places)

Now, we can calculate the change in velocity:

Change in velocity (v - u) = 0 m/s - 15 m/s
Change in velocity = -15 m/s

Next, we can calculate the average force exerted by the seatbelts using the formula mentioned earlier:

Force (F) = (mass x change in velocity) / time

F = (90 kg x -15 m/s) / 0.27 s
F = -5,000 N (rounded to the nearest whole number)

Since force is a vector quantity, the negative sign indicates that the force is exerted in the opposite direction to motion, which indicates a stopping force.

Therefore, the average force exerted by the seatbelts on the 90 kg passenger as the car is stopped is approximately 5,000 Newtons.

To calculate the average force exerted by the seatbelts on a passenger, we need to use the equation for average force:

Average force = Change in momentum / Time

First, we need to find the change in momentum. The momentum of an object is given by the formula:

Momentum = mass × velocity

In this case, the mass of the passenger is given as 90 kg, and the initial velocity is 15 m/s. The final velocity is 0 m/s since the car is brought to a rest. Therefore, the change in momentum is:

Change in momentum = (mass × final velocity) - (mass × initial velocity)
= mass × (final velocity - initial velocity)
= 90 kg × (0 m/s - 15 m/s)
= -1350 kg·m/s

Next, we need to find the time it takes for the car to stop. To do this, we can use the following equation:

Distance = (initial velocity × time) + (0.5 × acceleration × time^2)

In this case, the distance is given as 2.0 m, the initial velocity is 15 m/s, and the final velocity is 0 m/s. The acceleration can be found using the formula:

Acceleration = (final velocity - initial velocity) / time

Since the final velocity is 0 m/s, the acceleration becomes:

Acceleration = (0 m/s - 15 m/s) / time
= -15 m/s / time

Now, we can substitute this acceleration back into the equation for distance:

2.0 m = (15 m/s × time) + (0.5 × (-15 m/s / time) × time^2)

Simplifying this equation,

2.0 m = 15 m/s × time - 7.5 m/s × time
2.0 m = 7.5 m/s × time

Now, solve for time:

time = 2.0 m / 7.5 m/s
time ≈ 0.267 s

Finally, we can calculate the average force by plugging the values into the formula:

Average force = Change in momentum / Time
Average force = -1350 kg·m/s / 0.267 s
Average force ≈ -5056.18 N

The negative sign indicates that the force is in the opposite direction of the motion, which is expected since the car is brought to a rest. Therefore, the average force exerted by the seatbelts on the passenger is approximately 5056.18 N.